Question

我遍历文件并希望构建一个多级动态字典。最后一级需要存储值列表。

myDict = defaultdict(dict)

for key in lvlOneKeys: # this I know ahead of time so I set up my dictionary with first level. I'm not sure if this is necessary.
    myDict[key] = {}

with open(myFile, "rb") as fh:
    for line in fh:
        # found something, which will match lvlOneKey and dynamically determine lvlTwoKey and valueFound
        # ...
        myDict[lvlOneKey][lvlTwoKey].append(valueFound)

我需要这样做，因为lvlTwoKey会在不同的valueFound中被多次找到。

不幸的是，此代码导致lvlOneKey的KeyError。我做错了什么？

Answer 1

这是一种确保您不会收到错误的几乎万无一失的方法。我们定义myDict的方式，您可以拥有字典“级别1”和字典“级别2”的任何键。默认情况下，在字典树的末尾假定一个空列表。

myDict = defaultdict(lambda: defaultdict(list))

with open(myFile, "rb") as fh:
    for line in fh:
        # found something, which will match lvlOneKey and dynamically determine lvlTwoKey and valueFound
        # ...
        myDict[lvlOneKey][lvlTwoKey].append(valueFound)

Answer 2

使用以下内容替换for循环下的代码可以解决您的问题：

# if there is no `lvlTwoKey` in `myDict`, initialize a list with `valueFound` in it  
if not myDict[lvlOneKey].get(lvlTwoKey, None):
    myDict[lvlOneKey][lvlTwoKey] = [valueFound]
# otherwise, append to the existing list 
else: 
    myDict[lvlOneKey][lvlTwoKey].append(valueFound)

这使用字典上的get()方法，您可以阅读here。除此之外，它只是一个标准字典，我发现它通常比复杂的默认字典更具可读性/直观性。

Answer 3

如果进一步嵌套，这会有点烦人，但使用这种模式会有效。

l1keys = [1, 2, 1]
l2keys = ['foo', 'bar', 'spangle']

base = {}
for l1key, l2key in zip(l1keys, l2keys):
    for i in range(5):
        l1 = base.get(l1key, {})
        l2 = l1.get(l2key, [])
        l2.append(i)
        l1[l2key] = l2
        base[l1key] = l1

assert base == {
    1: {'foo': [0, 1, 2, 3, 4], 'spangle': [0, 1, 2, 3, 4]},
    2: {'bar': [0, 1, 2, 3, 4]}
}

如何在多级字典中附加到列表？

3 个答案: